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A line has a slope of −2-2 and includes the points (−3,j)(-3,j) and (−6,6)(-6,6). What is the value of jj?\newlinej=___j = \_\_\_

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Full solution

Q. A line has a slope of −2-2 and includes the points (−3,j)(-3,j) and (−6,6)(-6,6). What is the value of jj?\newlinej=___j = \_\_\_
  1. Identify Slope Formula: To find the value of jj, we can use the slope formula, which is (y2−y1)/(x2−x1)=slope(y_2 - y_1) / (x_2 - x_1) = \text{slope}, where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of two points on the line.
  2. Assign Coordinates: We know the slope mm is −2-2, and we have the points (−3,j)(-3, j) and (−6,6)(-6, 6). Let's assign (−3,j)(-3, j) as (x1,y1)(x_1, y_1) and (−6,6)(-6, 6) as (x2,y2)(x_2, y_2).
  3. Apply Slope Formula: Now we plug the values into the slope formula: (6−j)/(−6−(−3))=−2(6 - j) / (-6 - (-3)) = -2.
  4. Simplify Denominator: Simplify the denominator: −6+3=−3-6 + 3 = -3, so we have (6−j)/−3=−2(6 - j) / -3 = -2.
  5. Solve for jj: To find the value of jj, we need to solve for jj in the equation 6−j−3=−2\frac{6 - j}{-3} = -2. We can start by multiplying both sides of the equation by −3-3 to get rid of the denominator: −3×(6−j−3)=−3×−2-3 \times \left(\frac{6 - j}{-3}\right) = -3 \times -2.
  6. Solve for j: To find the value of jj, we need to solve for jj in the equation 6−j−3=−2\frac{6 - j}{-3} = -2. We can start by multiplying both sides of the equation by −3-3 to get rid of the denominator: −3×(6−j−3)=−3×−2-3 \times \left(\frac{6 - j}{-3}\right) = -3 \times -2.This simplifies to 6−j=66 - j = 6.

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